feat(work): add solutions for LeetCode problems 217, 242, 189, and 49

work/Easy/Array/217.contains-duplicate.js

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+/*
+ * 217. Contains Duplicate
+ * Difficulty: Easy
+ * https://leetcode.com/problems/contains-duplicate/
+ * 
+ * ──────────────────────────────────────────────────
+ * 
+ * Given an integer array nums, return true if any value appears at
+ * least twice in the array, and return false if every element is
+ * distinct.
+ * 
+ *  
+ * 
+ * Example 1:
+ * 
+ * Input: nums = [1,2,3,1]
+ * 
+ * Output: true
+ * 
+ * Explanation:
+ * 
+ * The element 1 occurs at the indices 0 and 3.
+ * 
+ * Example 2:
+ * 
+ * Input: nums = [1,2,3,4]
+ * 
+ * Output: false
+ * 
+ * Explanation:
+ * 
+ * All elements are distinct.
+ * 
+ * Example 3:
+ * 
+ * Input: nums = [1,1,1,3,3,4,3,2,4,2]
+ * 
+ * Output: true
+ * 
+ *  
+ * 
+ * Constraints:
+ * 
+ * 	• 1 <= nums.length <= 10^5
+ * 
+ * 	• -10^9 <= nums[i] <= 10^9
+*/
+
+/**
+ * @param {number[]} nums
+ * @return {boolean}
+ */
+var containsDuplicate = function(nums) {
+  const freq = {};
+  for (let n of nums) {
+    freq[n] = (freq[n] || 0) + 1;
+    if (freq[n] >= 2) {
+      return true;
+    }
+  } 
+  return false;
+};

work/Easy/Hash Table/242.valid-anagram.js

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+/*
+ * 242. Valid Anagram
+ * Difficulty: Easy
+ * https://leetcode.com/problems/valid-anagram/
+ * 
+ * ──────────────────────────────────────────────────
+ * 
+ * Given two strings s and t, return true if t is an anagram of s, and
+ * false otherwise.
+ * 
+ *  
+ * 
+ * Example 1:
+ * 
+ * Input: s = "anagram", t = "nagaram"
+ * 
+ * Output: true
+ * 
+ * Example 2:
+ * 
+ * Input: s = "rat", t = "car"
+ * 
+ * Output: false
+ * 
+ *  
+ * 
+ * Constraints:
+ * 
+ * 	• 1 <= s.length, t.length <= 5 * 10^4
+ * 
+ * 	• s and t consist of lowercase English letters.
+ * 
+ *  
+ * 
+ * Follow up: What if the inputs contain Unicode characters? How would
+ * you adapt your solution to such a case?
+*/
+
+/**
+ * @param {string} s
+ * @param {string} t
+ * @return {boolean}
+ */
+var isAnagram = function(s, t) {
+  let freq = {};
+  for (let c of s) freq[c] = (freq[c] || 0) + 1;
+  
+  for (let c of t) {
+    if(!freq[c] || freq[c] === 0){
+      return false;
+    }
+    freq[c] = freq[c] - 1;
+  }
+  return true;
+};